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I'm asked to prove that $$\int_\mathcal{C}P(z)\,d\overline{z}=-2\pi i R^2P'(a)$$where $\mathcal{C}$ is the circle $|z-a|=R$ and $P(z)$ is a polynomial.

What I've done is to use the formula $$f^{(n)}(z)=\int_C\frac{f(\zeta)}{(\zeta-z)^{n+1}}\,d\zeta\,.$$ From here, I parametrized the circle, setting $z=a+Re^{i\theta}$, did some simple algebra, and got down to $$-2\pi i R^2 P'(a)=\int_0^{2\pi}P(a+Re^{i\theta})(-iRe^{-i\theta})\,d\theta.$$Now, what I want to do is say $-iRe^{-i\theta}=\overline{z'}$ so that $$-2\pi i R^2P'(a)=\int_\mathcal{C} P(z)\,d\overline{z},$$but I don't think this is valid, or if it is valid, I'm not really sure why that is the case. Any helpful pointers if it is invalid? If it's valid, will someone demystify the mystery for me about why it is true? Thanks.

Edit: So I'm thinking it has something to do with the following idea: if $z(\theta)=a+Re^{i\theta}$, then $$\frac{d}{d\theta}(\overline{a+Re^{i\theta}})\,d\theta=d\overline{z}.$$Am I on the right track?

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1 Answer 1

It is valid.

$$z=a + R e^{i\theta}$$

thus

$$z'=i\theta R e^{i\theta}$$

hence

$$\overline{z}'=\overline{z'}=\overline{i\theta R e^{i\theta}} = \overline{i}\overline{\theta}\overline{R} \overline{e^{i\theta}}=-i\theta R e^{\overline{i}\overline{\theta}}=-i\theta R e^{-i\theta}$$

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Everything after "It is valid." I had already accounted for, so I don't have any problems with that. Can you explain why it is valid? –  Clayton Feb 19 '13 at 7:35
    
Now I do not understand your question. What exactly is it that you do not understand? –  Udo Klein Feb 19 '13 at 7:41
    
Why does$$\int f(z(\theta))\overline{z}'(\theta)\,d\theta=\int f(z)\,d\overline{z}\quad?$$ –  Clayton Feb 19 '13 at 7:43
    
Almost by definition: en.wikipedia.org/wiki/Line_integral –  Udo Klein Feb 19 '13 at 7:51
    
@Clayton: use the fact that $(z-a)(\bar{z}-a) = R^2$. –  Ron Gordon Feb 19 '13 at 7:51

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